leetcode:Missing Number

最新推荐文章于 2022-05-05 16:50:30 发布

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本文介绍了一种解决寻找数组中缺失数字问题的高效算法，通过原地交换元素，实现O(n)时间复杂度和常数空间复杂度。代码示例详细解释了实现过程，特别适用于面试和日常编程场景。

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Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.

For example,
Given nums = [0, 1, 3] return 2.

Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

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class Solution {
    
public:
    int missingNumber(vector<int>& nums) {
        
        int found = false;
        int retValue;
        for (int i=0; i<nums.size(); i++)
        {
            while (nums[i] != i)
            {
                int nextIdx = nums[i];
                
                if (nextIdx < nums.size())
                {
                    int tmp = nums[i];
                    nums[i] = nums[nextIdx];
                    nums[nextIdx] = tmp;
                }
                else
                {
                    found = true;
                    retValue = i; //注意，这里返回的是i不是nums[i];
                    break;
                }
            }
        }
        
        if (!found)
            return nums.size();
        else
            return retValue;
    }
};